$f(n) = 7n^{2}-3n-3-h(n)$ $h(t) = -t^{2}+t+3$ $ f(h(3)) = {?} $
Explanation: First, let's solve for the value of the inner function, $h(3)$ . Then we'll know what to plug into the outer function. $h(3) = -3^{2}+3+3$ $h(3) = -3$ Now we know that $h(3) = -3$ . Let's solve for $f(h(3))$ , which is $f(-3)$ $f(-3) = 7(-3)^{2}+(-3)(-3)-3-h(-3)$ To solve for the value of $f$ , we need to solve for the value of $h(-3)$ $h(-3) = -(-3)^{2}-3+3$ $h(-3) = -9$ That means $f(-3) = 7(-3)^{2}+(-3)(-3)-3-(-9)$ $f(-3) = 78$